What is the integral of (at)/(bt^2+c) ?

The integral of (at)/(bt^2+c) is a 1/(2b)ln|c+bt^2|+C

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integral of (at)/(bt^2+c)

\int \frac{a t}{b t ^{2} + c} d t

a \frac{1}{2 b} ln c + b t^{2} + C

Solution steps

\int \frac{a t}{b t ^{2} + c} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= a \cdot \int \frac{t}{c + b t ^{2}} d t

Apply u-substitution

= a \cdot \int \frac{1}{2 b u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= a \frac{1}{2 b} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= a \frac{1}{2 b} ln ∣ u ∣

Substitute back

u = c + b t^{2}

= a \frac{1}{2 b} ln c + b t^{2}

Add a constant to the solution

= a \frac{1}{2 b} ln c + b t^{2} + C

What is the integral of (at)/(bt^2+c) ?
The integral of (at)/(bt^2+c) is a 1/(2b)ln|c+bt^2|+C